I am having hard time in calculating (or constructing) $\displaystyle\frac{\mathbb C[x,y]}{\langle y^2 - x^3 - x\rangle}$.
I tried homogenizing the ideal $y^2 - x^3 -x $ to $ wy^2 - x^3 - xw^2$. But from here how to define a morphism into a nice ring such that the ideal goes to kernel.
The answer is negative. Assume the contrary and consider $f,g\in\mathbb C[t]$ that correspond to (the residue classes of) $x,y$ in our ring. Then $g^2=f^3+f$. Since $f$ and $g$ can't be both constants we have $\deg f\ge1$ and $\deg g\ge 1$. Let $p\mid f$ be an irreducible polynomial. Write $f=pf_1$ and (necessarily) $g=pg_1$, and thus we get $pg_1^2=f_1(p^2f_1^2+1)$. It follows that $p\mid f_1$, so $f_1=pf_2$ and plugging this into the last equation we get $g_1^2=f_2(p^4f_2^2+1)$. Continuing this way we arrive to an equation of the form $v^2=c(c^2u^4+1)$ with $c\in\mathbb C$, $c\ne0$, and $u,v\in\mathbb C[t]$ of degree $\ge 1$. This is equivalent to $(v-c\sqrt cu^2)(v+c\sqrt cu^2)=c$ and therefore $v\pm c\sqrt cu^2$ are constants, hence $u,v$ are constants, a contradiction.